First, recall the following identity: $$tan(\sum_{k=1}^{n}x_k)= i\cdot \frac{A-B}{A+B} \ ;\ A=\prod_{k=1}^{n}(1-i\cdot tan(x_k)) \ ,\ B=\prod_{k=1}^{n}(1+i\cdot tan(x_k))$$ Now, suppose the contrary, take $tan$ from both sides, using the aforementioned identity we conclude that for all $N$, there exist $A,B$ such that $$N=i\cdot \frac{A-B}{A+B}$$where $A=\prod_{k=1}^{2020} (1-ik)^{a_k}$ and $B=\prod_{k=1}^{2020} (1+ik)^{a_k}$. Note that $A,B\in \mathbb{Z}[i]$ and define $M=\prod_{k=1}^{2020} (1+k^2)$. from now on, we work in $\mathbb{Z}[i]$, it is clear that $rad(A)\mid M$ now, by the Chinese remainder theorem, there exists $N\in \mathbb{N}$ such that $gcd(N^2+1,2M)=1$. now $$N=i\cdot \frac{A-B}{A+B} \Rightarrow N+i=\frac{2iA}{A+B}$$so $\frac{2iA}{A+B}\in \mathbb{Z}[i]$ so now for a prime $p\in \mathbb{Z}[i]$ if $p\mid \frac{2iA}{A+B}$ then $p\mid rad(2A) \mid 2M$ and $p\mid N+i\mid N^2+1$ but $gcd(N^2+1,2M)=1$, a contradiction.